Question: Let $h$ be a polynomial function and let $h'$, its derivative, be defined as $h'(x)=x^2(x-2)^2(x-1)^2 $. At how many points does the graph of $h$ have a relative maximum ? Choose 1 answer: Choose 1 answer: (Choice A) A None (Choice B) B One (Choice C) C Two (Choice D) D Three
We can find the relative extrema (i.e. minima and maxima) of $h$ by looking for the intervals where its derivative $h'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $h'(x)=x^2(x-2)^2(x-1)^2$. $h'(x)=0$ for $x=0,1,2$. Since $h'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=0$, $x=1$, and $x=2$. $h$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into four intervals: $\llap{-}1$ $0$ $1$ $2$ $3$ $x< 0$ $0<x<1$ $1<x<2$ $x>2$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $x<0$ $x=-1$ $h'(-1)=36>0$ $h$ is increasing $\nearrow$ $0<x<1$ $x=\dfrac{1}{2}$ $h'\left(\dfrac12\right)=\dfrac{9}{64}>0$ $h$ is increasing $\nearrow$ $1<x<2$ $x=\dfrac{3}{2}$ $h'\left(\dfrac32\right)=\dfrac{9}{64}>0$ $h$ is increasing $\nearrow$ $x>2$ $x=3$ $h'(3)=36>0$ $h$ is increasing $\nearrow$ Now let's look at the critical points: $x$ Before After Verdict $0$ $\nearrow$ $\nearrow$ Not an extremum $1$ $\nearrow$ $\nearrow$ Not an extremum $2$ $\nearrow$ $\nearrow$ Not an extremum Now we can see that $h$ has no relative maxima.